Linear combinations of random variables

6.2 Linear combinations of random variables

One variable

Adding a constant moves the distribution but does not spread it. Multiplying by a constant does both.

$\mathrm{E}(3X - 2) = 13$ and $\mathrm{Var}(3X - 2) = 9 \times 4 = 36$.

The constant $-2$ makes no difference to the variance. A standard deviation scales by $\lvert a\rvert$, which makes the standard deviation here $3 \times 2 = 6$.

Two variables

Variances add even for a difference, because squaring $b = -1$ gives $+1$:

$\mathrm{Var}(X - Y) = \mathrm{Var}(X) + \mathrm{Var}(Y)$

A difference between two uncertain quantities has more uncertainty than either one, not less.

Combining normal variables

Let $D = Y - 2X$. Then

$\mathrm{E}(D) = 25 - 20 = 5 \qquad \mathrm{Var}(D) = 9 + 4 \times 4 = 25$

So $D \sim \mathrm{N}(5, 25)$, and

$\mathrm{P}(D > 0) = \mathrm{P}\left(Z > \frac{0 - 5}{5}\right) = \mathrm{P}(Z > -1) = 0.8413$

$X_1 + X_2$ and $2X$

These look the same but are not.

Four different apples means $X_1 + X_2 + X_3 + X_4$, so the mean is $4 \times 150 + 200 = 800$ and the variance is $4 \times 100 = 400$. The box's fixed mass adds nothing to the variance.

$\mathrm{P}(T > 850) = \mathrm{P}\left(Z > \frac{850 - 800}{20}\right) = 1 - \Phi(2.5) = 0.0062$

$D = Y - 2X$ has mean $320 - 300 = 20$ and variance $400 + 4 \times 100 = 800$.

$\mathrm{P}(D > 0) = \mathrm{P}\left(Z > \frac{-20}{\sqrt{800}}\right) = \Phi(0.707) = 0.760$

Combining Poisson variables

$X + Y \sim \mathrm{Po}(5)$, so

$\mathrm{P}(2) = \mathrm{e}^{-5}\frac{5^2}{2!} = 0.0842$

Which results need independence